Problem: Is ${279126}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {279126}= &&{2}\cdot100000+ \\&&{7}\cdot10000+ \\&&{9}\cdot1000+ \\&&{1}\cdot100+ \\&&{2}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {279126}= &&{2}(99999+1)+ \\&&{7}(9999+1)+ \\&&{9}(999+1)+ \\&&{1}(99+1)+ \\&&{2}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {279126}= &&\gray{2\cdot99999}+ \\&&\gray{7\cdot9999}+ \\&&\gray{9\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{2\cdot9}+ \\&& {2}+{7}+{9}+{1}+{2}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${279126}$ is divisible by $9$ if ${ 2}+{7}+{9}+{1}+{2}+{6}$ is divisible by $9$ Add the digits of ${279126}$ $ {2}+{7}+{9}+{1}+{2}+{6} = {27} $ If ${27}$ is divisible by $9$ , then ${279126}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${279126}$ must also be divisible by $9$.